Problem: Let $f(x)=x\ln(x)$. What is the absolute minimum value of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{e}$ (Choice C) C $-\dfrac{1}{e}$ (Choice D) D $f$ has no minimum value
Let's first find the relative extremum points of $f$, and then consider them along with the function's end behavior in both directions. Note that the domain of $f$ is all real numbers such that $x>0$. We start with finding the critical points of $f$. The derivative of $f$ is $f'(x)=\ln(x)+1$. $f'(x)=0$ for $x=\dfrac{1}{e}$. $f'$ is defined for all real numbers in the domain of $f$. Therefore, our only critical point is $x=\dfrac{1}{e}$. Our critical point divides the function's domain $($ $x>0$ $)$ into two intervals: $0$ $1$ $2$ $3$ $0<x<\frac{1}{e}$ $x>\frac{1}{e}$ $\frac{1}{e}$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $0<x<\dfrac{1}{e}$ $x=\dfrac{1}{e^2}$ $f'\left(\dfrac{1}{e^2}\right)=-1<0$ $f$ is decreasing $\searrow$ $x>\dfrac{1}{e}$ $x=1$ $f'(1)=1>0$ $f$ is increasing $\nearrow$ Let's imagine ourselves walking on the graph of $f$, starting at the left end of the domain $($ which is $0$ $)$ and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until we reach $x=\dfrac{1}{e}$. Then, we will be forever going up. Therefore, $f$ must obtain its absolute minimum value at $x=\dfrac{1}{e}$. We are asked to find that minimum value, which is $f\left(\dfrac{1}{e}\right)=-\dfrac{1}{e}$. In conclusion, the absolute minimum value of $f$ is $-\dfrac{1}{e}$.